The Completion of a Measure Space

The Completion of a Measure Space

Recall from the Complete Measure Spaces page that a measure space $(X, \mathcal, \mu)$ is said to be complete if for every measurable set $E \in \mathcal P(X)$ with $\mu(E) = 0$ we have that every subset of $E$ is a measurable set.

Given an incomplete measure space $(X, \mathcal A, \mu)$ we can construct a minimal "larger" measure space $(X, \overline{\mathcal A}, \overline{\mu})$ that is a complete measure space where $\mathcal A \subseteq \overline{\mathcal A}$, and $\overline{\mu} : \overline{\mathcal A} \to [0, \infty]$ is an extension measure of $\mu$ on $\overline{\mathcal A}$.

 Theorem 1: Let $(X, \mathcal A, \mu)$ be a measure space. Then there exists a minimmal measure space $(X, \overline{\mathcal A}, \overline{\mu})$ with $\mathcal A \subseteq \overline{\mathcal A}$ and $\overline{\mu}$ is an extension of $\mu$ on $\overline{\mathcal A}$ where $\overline{\mathcal A} = \{ A \cup E : A \in \mathcal A, \: E \subseteq F \: \mathrm{for \: some \:} F \in \mathcal A \: \mathrm{with} \: \mu (F) = 0 \}$, and $\overline{\mu} : \overline{\mathcal A} \to [0, \infty]$ is defined for all $A \cup E \in \overline{\mathcal A}$ by $\overline{\mu}(A \cup E) = \mu(A)$, with: a) $\overline{\mathcal A}$ is a $\sigma$-algebra. b) $\overline{\mu}$ is a well-defined function. c) $\overline{\mu}$ is a measure on $\overline{\mathcal A}$. d) $(X, \overline{\mathcal A}, \overline{\mu})$ is a complete measure space. e) $(X, \overline{\mathcal A}, \overline{\mu})$ is a minimal extended measure space of $(X, \mathcal A, \mu)$ that is complete.
• Proof of a) Let $X \in \overline{\mathcal A}$. Then $X = A \cup E$ where $A \in \mathcal A$ and $E \subseteq F$ for some $F \in \mathcal A$ with $\mu (F) = 0$. Then:
(1)
\begin{align} \quad X^c &= (A \cup E)^c \\ &= (A \cup (F \setminus (F \setminus E)))^c \\ &= (A \cup (F \cap (F \setminus E)^c))^c \\ &= A^c \cap [F \cap (F \setminus E)^c]^c \\ &= A^c \cap [F^c \cup (F \setminus E)] \\ &= (A^c \cap F^c) \cup (A^c \cap (F \setminus E)) \end{align}
• Since $A, F \in \mathcal A$ we have that $(A^c \cap F^c) \in \mathcal A$. And since $A^c \cap (F \setminus E) \subseteq F$ and $\mu(F) = 0$, we conclude that $X^c \in \overline{\mathcal A}$. So $\overline{\mathcal A}$ is closed under complementation.
• Let $(X_n)_{n=1}^{\infty} \subseteq \overline{\mathcal A}$. Then for each $n \in \mathbb{N}$ we have that $X_n = A_n \cup E_n$ where $A_n \in \mathcal A$ and $E_n \subseteq F_n$ for some $F_n \in \mathcal A$ with $\mu (F_n) = 0$. Then:
(2)
\begin{align} \quad \bigcup_{n=1}^{\infty} X_n &= \bigcup_{n=1}^{\infty} [A_n \cup E_n] \\ &= \left ( \bigcup_{n=1}^{\infty} A_n \right ) \cup \left ( \bigcup_{n=1}^{\infty} E_n \right ) \end{align}
• Since $A_n \in \mathcal A$ for each $n \in \mathbb{N}$, we have that $\displaystyle{\bigcup_{n=1}^{\infty} A_n \in \mathcal A}$. Furthermore, $\displaystyle{\left ( \bigcup_{n=1}^{\infty} E_n \right ) \subseteq \left ( \bigcup_{n=1}^{\infty} F_n \right )}$ where $\displaystyle{\left ( \bigcup_{n=1}^{\infty} F_n \right ) \in \mathcal A}$ and $\displaystyle{\mu \left ( \bigcup_{n=1}^{\infty} E_n \right ) = 0}$ since $(X, \mathcal A, \mu)$ is a measure space. Therefore $\displaystyle{\bigcup_{n=1}^{\infty} X_n \in \overline{\mathcal A}}$. So $\overline{\mathcal A}$ is a $\sigma$-algebra on $X$.
• Proof of b) Let $X \in \overline{\mathcal A}$ be such that $X = A \cup E$ and $X = A' \cup E'$ where $A, A' \in \mathcal A$ and $E \subseteq F$, $E' \subseteq F'$ where $F, F' \in \mathcal A$ and $\mu (F) = 0$ and $\mu (F') = 0$. Then:
(3)
\begin{align} \quad (A \cup E) \cup (F \cup F') = (A' \cup E') \cup (F \cup F') \quad \Leftrightarrow \quad A \cup F \cup F' = A' \cup F \cup F' \end{align}
• Therefore:
(4)
\begin{align} \quad \mu (A) \leq \mu(A \cup F \cup F') = \mu (A' \cup F \cup F') \leq \mu (A') + \mu (F \cup F') = \mu (A') + 0 = \mu(A') \end{align}
(5)
\begin{align} \quad \mu (A') \leq \mu (A' \cup F \cup F') = \mu(A \cup F \cup F') \leq \mu (A) + \mu (F \cup F') = \mu (A) + 0 = \mu (A) \end{align}
• Hence $\mu (A) = \mu(A')$. So $\overline{\mu} (A \cup E) = \overline{\mu} (A' \cup E')$ and $\overline{\mu}$ is well-defined.
• Proof of c) For each $A \in \mathcal A$ we have that $A = A \cup \emptyset$, so $\overline{\mu} (A) = \mu (A)$. So $\overline{\mu}$ is an extension of $\mu$ on $\mathcal A$.
• Furthermore we have that:
(6)
\begin{align} \quad \mu(\emptyset) = \mu(\emptyset) = 0 \end{align}
• Let $(X_n)_{n=1}^{\infty}$ be a countable collection of mutually disjoint sets in $\overline{\mathcal A}$. Then for each $n \in \mathbb{N}$ we have that $X_n = A_n \cup E_n$ where $A_n \in \mathcal A$ and $E_n \subseteq F_n$ for some $F_n \in \mathcal A$ with $\mu(F_n) = 0$. Now:
(7)
\begin{align} \quad \overline{\mu} \left ( \bigcup_{n=1}^{\infty} X_n \right ) = \mu \left ( \bigcup_{n=1}^{\infty} A_n \right ) = \sum_{n=1}^{\infty} \mu (A_n) = \sum_{n=1}^{\infty} \overline{\mu} (X_n) \end{align}
• So $\overline{\mu}$ is a measure on $\overline{\mathcal A}$ and $(X, \overline{\mathcal A}, \overline{\mu})$ is a measure space.
• Proof of d) Let $F \in \overline{\mathcal A}$ be such that $\overline{\mu} (F) = 0$ and let $E \subseteq F$. Since $F \in \overline{\mathcal A}$ we have that $F = A \cup E'$ for some $A \in \mathcal A$ and where $E' \subseteq F'$ for some $F' \in \mathcal A$ with $\mu (F') = 0$. Now:
(8)
\begin{align} \quad \overline{\mu (F)} = \mu(A) = 0 \end{align}
• Therefore $\mu (A \cup E') \leq \mu (A) \cup \mu (E') = 0 + 0$. So $E = \emptyset \cup E$ where $\emptyset \in \mathcal A$ and $E \subseteq F = (A \cup E')$ with $\mu(F) = 0$. So $E \in \overline{\mathcal A}$. Thus $(X, \overline{\mathcal A}, \overline{\mu})$ is a complete measure space.
• Proof of e) Let $(X, \mathcal B, \eta)$ be a $\sigma$-algebra with $\mathcal A \subseteq \mathcal B$ and with $\eta$ an extension measure of $\mu$ to $\mathcal B$, such that $(X, \mathcal B, \eta)$ is a complete measure space. Let $X \in \overline{\mathcal A}$. Then $X = A \cup E$ where $A \in \mathcal A$ and $E \subseteq F$ for some $F \in \mathcal A$ with $\mu (F) = 0$. Then $\eta (F) = 0$. Since $E \subseteq F$ and $(X, \mathcal B, \eta)$ is complete we have that $E \in \mathcal B$. So $X = A \cup E \in \mathcal B$. Therefore $\overline{\mathcal A} \subseteq \mathcal B$, which shows tat $(X, \overline{\mathcal A}, \overline{\mu})$ is minimal. $\blacksquare$

The measure space defined above is given a special name.

 Definition: Let $(X, \mathcal A, \mu)$ be a measure space. Then $(X, \overline{\mathcal A}, \overline{\mu})$ is the Completion of $(X, \mathcal A, \mu)$.